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Theorem 6.19. In short, we want to show

$$\left\|\int |f(\cdot, y)|\,d\nu(y)\right\|_p \leq \int \|f(\cdot, y)\|_p \, d\nu(y)$$

for m'ble $f$.


Could we prove this using just the duality of the norm? That is, there exists some $g \in L^q(\mu)$ such that

$$ \begin{align} \left\|\int |f(\cdot, y)|\,d\nu(y)\right\|_p &= \sup_{\|g\|_q = 1} \iint |f(x,y)||g(x)| \, d\mu(x) \, d\nu(y) \\ &= \sup_{\|g\|_q = 1}\iint |f(x,y)||g(x)| \, d\nu(y) \, d\mu(x) \tag{Fubini} \\ &\leq \int \left(\sup_{\|g\|_q = 1} \int |f(x,y)| |g(x)| \, d\nu(y) \right) \, d\mu(x) \tag{$\dagger$}\\ &= \int \|f(\cdot, y)\|_p \, d\nu(y). \end{align} $$

where for $(\dagger)$, I used the fact that for m'ble $h$ we have, $\sup\int h \leq \int \sup h$.

Am I overlooking anything? It seems this is nicer than the approach used in the book.

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    $\begingroup$ This proof is fine. $\endgroup$ Feb 21, 2021 at 0:23

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